Question

O conjunto r em solução da equação exponencial (0,2)^x+1= √125, COMO FAZER?

Answer 1

Olá,

use as propriedades da exponenciação:

$\dfrac{1}{a1}=a^{-1}$

$(a^m)^n=a^{m\cdot n$

$\sqrt[n]{a}= \sqrt[n]{a^1}=a^{ \tfrac{1}{n}}$

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$(0,2)^{x+1}= \sqrt{125}\\\\ \left( \dfrac{2}{10}\right)^{x+1}= \sqrt[2]{5^3}\\\\ \left( \dfrac{2\div2}{10\div2}\right)^{x+1}=5^{ \tfrac{3}{2}}\\\\ \left( \dfrac{1}{5}\right)^{x+1}=5^{ \tfrac{3}{2}}\\\\ \left( \dfrac{1}{5^1}\right)^{x+1}=5^{ \tfrac{3}{2}}\\\\ (5^{-1})^{x+1}=5^{ \tfrac{3}{2}}\\ \not5^{-x-1}=\not5^{ \tfrac{3}{2}}\\\\ -x-1= \dfrac{3}{2}\\\\ -x=1+ \dfrac{3}{2}\\\\ -x= \dfrac{1\cdot2+3}{2}\\\\ -x= \dfrac{5}{2}~~(multiplique~por~-1)\\\\ x=- \dfrac{5}{2}$

$\Large\boxed{\text{S}=\left\{- \dfrac{5}{2}\right\}}$

Tenha ótimos estudos ;P